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Is there a 'nice' proof to show that $\det(E^T) = \det(E)$ where $E$ is an elementary matrix?

Clearly it's true for the elementary matrix representing a row being multiplied by a constant, because then $E^T = E$ as it is diagonal.

I was thinking for the "row-addition" type, it's clearly true because if $E_1$ is a matrix representing row-addition then it is either an upper/lower triangular matrix and so $\det(E_1)$ is equal to the product of the diagonals. If $E_1$ is an upper/lower triangular matrix, then $E_1^T$ is a lower/upper triangular matrix and so $\det(E_1^T) = \det(E_1)$ as the diagonal entries remain the same when the matrix is transposed.

How about for the "row-switching" matrix where rows $i$ and $j$ have been swapped on the identity matrix? Can we use the linearity of the rows in a matrix somehow?

Thanks for any help!

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It is not true that $\det{M} = \mathrm{tr}\,M$ for upper or lower triangular matrices. –  josh Mar 20 '13 at 1:35
    
@josh Sorry, I've been massively stupid and thought the trace was the product of the diagonals NOT the sum. –  Noble. Mar 20 '13 at 1:37
    
@josh I've edited the main post, sorry for the confusion! –  Noble. Mar 20 '13 at 1:39
    
no big deal, glad you understand more now too. –  josh Apr 2 '13 at 2:28

1 Answer 1

up vote 2 down vote accepted

You can use the fact that switching two rows or columns of a matrix changes the sign of the determinant. Switching two rows of $E$ makes it diagonal, then switch the corresponding columns and you have $E^T$

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Ah right, yes. Also, because you've switched rows/columns twice you've essentially multiplied the determinant by $(-1)^2$ and hence it's the same! Many thanks. –  Noble. Mar 20 '13 at 1:54

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